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You are taking a shortcut when you say, "The voltage is zero." Voltage is always measured between two points. In electrical engineering, when we say the voltage at point X is V, we actually are measuring the voltage between point X and an implicit other point called "ground". In the electric power grid, "neutral" is ground, by definition. So the voltage ...


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How quickly discharge will occur in the situation you sketch depends entirely on the surface properties of the negative electrode. For current to flow, electrons need to be released from the negative electrode; once they are free, they will accelerate unimpeded to the positive electrode. They will arrive there with 1.5 eV of energy, causing a small amount of ...


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Like all steady state circuit analysis things in parallel share the same voltage, and things in series share the same current. The question then really boils down to what the current vs. voltage curve looks like for a solar panel. A quick search turned up this: from https://www.folsomlabs.com/modeling/module/module_model So in this case it looks like you ...


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for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


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No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.


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NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.


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In vacuum there are no charge carriers like ions or electrons. With nothing to carry charge, i.e. current, such a battery would discharge much, much slower than when the battery poles are connected by something that can carry charge like a conductor or an imperfect insulator.


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Lines of electrostatic force exist between the positive and negative poles of the battery, even though they're separated by a vacuum. Vacuum permittivity is ε0 = 8.854 * 10^-12 farads per meter. By convention, this is called the dielectric constant of 1, a baseline against which the dielectric permittivities of other materials are compared. ...


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Electrons (and other charge carriers, e.g., ions) in vacuum travel without resistance. However, as pointed out, correctly, in the other answers, there are no charge carriers in vacuum. Nevertheless, electrons can escape from the terminals if they have a kinetic energy which is bigger than the potential barrier of the terminal surface, i.e., the work ...


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I think this picture answers the question There battery's terminals are not charged. It's just a chemical reaction that starts the charge. There's a Zn and a CuSO4. When they meet each other via a cable, the Zn gives 2 electrons to ChSO4. Cu gets rid of SO4, leaves 2 extra electrons to SO4 and takes the 2 electrons it got from Zn. Then we have SO4 ...


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The voltage of a cell is caused by a chemical process, and the details of this process determine the behavior of the cell when a current flows. Simplistically, that behavior can be described as the cell having an internal (series) resistance - although the real behavior is more complex (an internal resistance would not explain that a battery goes flat after ...


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Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ...


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You have to think more carefully about what exactly $Q$,$I$ and $t$ signify: In $$ Q = I t$$ $Q$ is the charge that is transported by the current $I$ during the time $t$. If you now write $$ \frac{I}{Q} = \frac{1}{t}$$ then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...


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My brother-in-law faced this EXACT problem, as he worked on high tension lines. There is a corona discharge from these lines due to the very high voltages involved. From experience, the linemen learned that this corona discharge is injurious to internal organs. To prevent injury, the linemen wear the Faraday cage suit, because such a suit keeps the ...


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Well, yes you can, but it is usually very hard. Here are the steps: Solve the Laplace equation: $$ \nabla^2V = 0 \, .$$ In your case, find the general solution in spherical coordinates. Try to use every simplification you can. You might wonder why you don't solve Poisson's equation: $$ \epsilon\nabla^2V = \rho \, .$$ That's because a conductor is an equal ...



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